Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I recently came across this question:

Is the axiom of choice needed to prove the following statement:

Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. There exists a linear map $T : V \rightarrow W$ such that $Tv = w$.

I have talked to someone, who went and asked a few people, and they think that just ZF by itself is not sufficient, and that we indeed do need AC. Furthermore, they think that the statement is itself not sufficient to prove AC. Can anyone give a definitive answer?

share|improve this question
4  
It is consistent with ZF that there is a vector space whose dual space is zero, so you indeed need some form of choice. See mathoverflow.net/questions/49388/… –  Mariano Suárez-Alvarez Mar 24 at 9:09

1 Answer 1

Yes, the axiom of choice is needed.

Läuchli has constructed a vector space whose only endomorphisms are scalar multiplication. In such vector space, if $v\neq \alpha w$ there is no such $T$. The paper is in German, and it uses mathematical terminology which took some time to understand.

Läuchli, H. "Auswahlaxiom in der Algebra", Comment. Math. Helv. 37 (1962-1963), 1–18.

In my masters thesis I showed (by extending Läuchli's argument) that in fact $\sf ZF+DC_\kappa$ cannot prove the existence of such $T$, for any given $\kappa$. You can find a summary of the result (in a very early edition) here: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

To my knowledge, the question whether or not the assertion that such $T$ always exists implies the axiom of choice is still open.

share|improve this answer
    
Thank you @Martin. –  Asaf Karagila Mar 24 at 15:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.